Is there a nice characterization for the complete metric subspaces of $\mathbb{Q}$ (with the usual metric)?

It seems like a such a subspace must have empty interior; if it contained an open interval then it would clearly contain a non-converging Cauchy sequence. But this isn't enough, because consider the subset $\left\{\frac{1}{n} : n \in \mathbb{N} \right\} \subseteq \mathbb{Q}$. This has empty interior but is not complete because it doesn't contain $0$. In other words, a complete subspace of $\mathbb{Q}$ must be closed (which I think is true about metric spaces in general).

So a complete subspace of $\mathbb{Q}$ must be closed and have empty interior, but are these two sufficient to imply completeness? If so, how do I prove this, and if not, what am I missing? I'm not necessarily looking for a complete answer, just some pointers in the right direction.

  • $\begingroup$ Ah right, the Cantor set. $\endgroup$ – Vikram Saraph Jun 2 '14 at 19:27
  • $\begingroup$ So is there some nice property $P$ so that $A \subseteq \mathbb{Q}$ is complete if and only if $A$ is closed, $A$ has empty interior, and $P$ is true of $A$? $\endgroup$ – Vikram Saraph Jun 2 '14 at 19:33
  • $\begingroup$ Yeah, still thinking about that. :) $\endgroup$ – Thomas Andrews Jun 2 '14 at 19:34

There is a complete characterization of all countable complete metric spaces. Ordinals.

More specifically, one can show that if $\alpha$ is a countable ordinal then the order topology on $\alpha$ is completely metrizable.

Of course that there are complete subspaces of $\Bbb Q$ which are not order isomorphic to an ordinal; however they are homeomorphic to an ordinal (once we choose the metric, and we forget about the order).

[In my previous post I wrote something which was false; that successor ordinals were the complete ones, however I neglected the fact that a sequence whose distances from the first element approach $\infty$ need not have a limit. Successor ordinals are the compact metric spaces, though. So a subset of $\Bbb Q$ is compact if and only if it is homeomorphic to a successor ordinal.]

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  • $\begingroup$ Thanks! Regarding your last point, are you saying that a subset of $\mathbb{Q}$ is complete if and only if it's compact, or do you mean something else? $\endgroup$ – Vikram Saraph Jun 3 '14 at 18:02
  • $\begingroup$ Hmmm, I notice that something is a wee bit off here. $\Bbb N$ is complete, but not a successor ordinal (nor compact). There might be a slightly more complicated answer afoot. I'll think about it and if I won't be able to find a decent solution I'll remove my answer. $\endgroup$ – Asaf Karagila Jun 3 '14 at 21:00
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    $\begingroup$ Vikram, I couldn't find an elegant solution. (The only thing I managed to think of is that there is a sequence of intervals $[a_n,b_n]$ such that $|a_n|,|b_n|\to\infty$ and $X\cap[a_n,b_n]$ is compact; what I wrote is also false, every countable ordinal is completely metrizable; the successor ordinals are the compact ones.) $\endgroup$ – Asaf Karagila Jun 4 '14 at 17:46
  • $\begingroup$ Right, I think this is much better now. Sorry for my previous mistake. :-) $\endgroup$ – Asaf Karagila Jun 5 '14 at 1:05

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